Prove that: $$\dfrac{\tan A + \sin A}{\tan A - \sin A} = \dfrac{\sec A + 1}{\sec A - 1}$$
$${\textbf{Step -1: Simplifying the LHS part by applying Relevant Trigonometric Identity}}{\textbf{.}}$$
$$\dfrac{{\tan A + \sin A}}{{\tan A - \sin A}}$$
$$ = \dfrac{{\dfrac{{\sin A}}{{\cos A}} + \sin A}}{{\dfrac{{\sin A}}{{\cos A}} - \sin A}}$$ $$\left[\because \mathbf {tan \ A = \dfrac{{sin \ A}}{{cos \ A}}} \right]$$
$$ = \dfrac{{\sin A\left( {\dfrac{1}{{\cos A}} + 1} \right)}}{{\sin A\left( {\dfrac{1}{{\cos A }}-1} \right)}}$$
$$ = \dfrac{{\sec A + 1}}{{\sec A - 1}}$$ $$\left [\because \mathbf {\dfrac{1}{{cos \ A}} \pm 1 = sec \ A \pm 1} \right]$$
$${\textbf{}}{\textbf{ Hence, LHS = RHS proved}}{\textbf{.}}$$